# SOLVING ALGEBRA PROBLEMS - LINKS TO THEIR STUDY, SOLVED EXAMPLES, EXERCISES ON LOGARITHMS

Please study

Logarithm Formulas before Solving Algebra Problems

if you have not already done so.

Having the knowledge of the Formulae

is a prerequisite here.

### Solved Example 1 of Solving Algebra Problems :

Find the value of

(i) log_{343} 49 (ii) log_{0.01} (0.0001) (iii) log_{(2√3)} (1728)

** Solution to Example 1 of Solving Algebra Problems : **

(i) Let A = log_{343} 49

We know 49 = 7 x 7 = 7^{2}; 343 = 49 x 7 = 7 x 7 x 7 = 7^{3}

∴ A = log_{343} 49 = log_{(73)} (7^{2})

We know, in log of a power to the base of another power (See Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.

∴ A = (2⁄3) log_{7} 7

But log of any number to the same Base is one (see Formula 4).

∴ A = (2⁄3) (1) = 2⁄3. Ans.

(ii) Let A = log_{0.01} (0.0001)

We know 0.0001 = 0.01 x 0.01 = (0.01)^{2};

∴ A = log_{(0.01)} (0.0001) = log_{(0.01)} {(0.01)^{2}}

We know, in log of a power (see Formula 7), the exponent multiplies the log.

∴ A = 2 log_{(0.01)} (0.01)

But log of any number to the same Base is one (see Formula 4).

∴ A = 2 (1) = 2. Ans.

(iii) Let A = log_{(2√3)} (1728)

We know 2√3 = 2 x (3)^{(1⁄2)} = (2^{2})^{(1⁄2)} x (3)^{(1⁄2)}

= 4^{(1⁄2)} x (3)^{(1⁄2)} = (4 x 3)^{(1⁄2)} = 12^{(1⁄2)}

1728 = 12 x 144 = 12 x 12 x 12 = 12^{3}

See how the base (2√3) and 1728 are made as powers of the same number 12.

∴ A = log_{(2√3)} (1728) = log_{{12(1⁄2)}} (12^{3})

We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.

∴ A = {3⁄(1⁄2)} log_{12} 12

But log of any number to the same Base is one (see Formula 4).

∴ A = (3 x 2) (1) = 6.

Thus log_{(2√3)} (1728) = 6. Ans.

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### Solved Example 2 of Solving Algebra Problems :

Solve the following Problem on Solving Algebra Problems :

If (2.3)^{x} = (0.23)^{y} = 1000, show that 1⁄**x** - 1⁄**y** = 1⁄3

** Solution to Example 2 of Solving Algebra Problems : **

By data (2.3)^{x} = 1000

By converting from Exponential Form to Logarithmic Form (See Formula 1), we get

**x** = log_{(2.3)} 1000

Taking reciprocals, we get

Reciprocal of L.H.S. = 1⁄**x**

Reciprocal of R.H.S. = 1⁄(log_{(2.3)} 1000)

We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)

∴ Reciprocal of R.H.S. = log_{1000} (2.3)

∴ 1⁄**x** = log_{1000} (2.3)

By data (0.23)^{y} = 1000

By converting from Exponential Form to Logarithmic Form (See Formula 1), we get

**y** = log_{(0.23)} 1000

Taking reciprocals, we get

Reciprocal of L.H.S. = 1⁄**y**

Reciprocal of R.H.S. = 1⁄(log_{(0.23)} 1000)

We know, the reciprocal of the log of a quantity to a base is equal to the log with the quantity and the base interchanged.(See Formula 10)

∴ Reciprocal of R.H.S. = log_{1000} (0.23)

∴ 1⁄**y** = log_{1000} (0.23)

To prove 1⁄**x** - 1⁄**y** = 1⁄3

L.H.S. = 1⁄**x** - 1⁄**y** = log_{1000} (2.3) - log_{1000} (0.23)

We know, log of a quotient can be written as the difference of the log of the numerator and denominator of the quotient (See Formula 6)

Considering the reverse, the difference of the logs of different quantitiescan be written as log of a quotient (the quantity whose log has negative sign, in the denominator) of the quantities.

∴ L.H.S. = log_{1000} {(2.3)⁄(0.23)} = log_{1000} (10)

= log_{103} (10^{1})

We know, in log of a power to the base of another power (see Formula 9),the quotient of the exponents (exponent of the base in the denominator) multiplies the log.

∴ L.H.S. = (1⁄3) log_{10} 10

We know log of a quantity to the same base is 1. (See Formula 4)

∴ L.H.S. = (1⁄3) (1) = 1⁄3 = R.H.S. (Proved.)

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### Logarithm Tables : Characteristic and Mantissa

To find the log value of a number using Tables

we need the knowledge of Characteristic and Mantissa

which is covered in
Logarithm Tables.

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### Exercise : Solving Algebra Problems

Solve the following problems on Solving Algebra Problems

- If (log
_{2} **a**)⁄4 = (log_{2} **b**)⁄6 = (log_{2} **c**)⁄(3**p**) and **a**^{3}.**b**^{2}.**c** = 1,

find the value of **p**

- If 2 log
_{x} **a** + log_{ax} **a** + 3 log_{a2x} **a** = 0,

find the value of **x**.

For Answers see at the bottom of the page.

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### Answers to Exercise : Solving Algebra Problems

Answers to Problems on Solving Algebra Problems :

(1) -8 (2) **a**^{-1⁄2}, **a**^{-4⁄3}